试验任务1
1
#include <stdio.h>
#define N 4
int main()
{
int x[N] = {1, 9, 8, 4};
int i;
int *p;
// 方式1：通过数组名和下标遍历输出数组元素
for (i = 0; i < N; ++i)
printf("%d", x[i]);
printf("\n");
// 方式2：通过指针变量遍历输出数组元素 （写法1）
for (p = x; p < x + N; ++p)
printf("%d", *p);
printf("\n");
// 方式2：通过指针变量遍历输出数组元素（写法2）
p = x;
for (i = 0; i < N; ++i)
printf("%d", *(p + i));
printf("\n");
// 方式2：通过指针变量遍历输出数组元素（写法3）
p = x;
for (i = 0; i < N; ++i)
printf("%d", p[i]);
printf("\n");
return 0;
}
2
#include <stdio.h>
int main()
{
int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
int i, j;
int *p; // 指针变量，存放int类型数据的地址
int(*q)[4]; // 指针变量，指向包含4个int型元素的一维数组
// 使用数组名、下标访问二维数组元素
for (i = 0; i < 2; ++i)
{
for (j = 0; j < 4; ++j)
printf("%d", x[i][j]);
printf("\n");
} // 使用指针变量p间接访问二维数组元素
for (p = &x[0][0], i = 0; p < &x[0][0] + 8; ++p, ++i)
{
printf("%d", *p);
if ((i + 1) % 4 == 0)
printf("\n");
}
// 使用指针变量q间接访问二维数组元素
for (q = x; q < x + 2; ++q)
{
for (j = 0; j < 4; ++j)
printf("%d", *(*q + j));
printf("\n");
}
return 0;
}
实验任务2
1
#include <stdio.h>
#include <string.h>
#define N 80
int main()
{
char s1[] = "Learning makes me happy";
char s2[] = "Learning makes me sleepy";
char tmp[N];
printf("sizeof(s1) vs. strlen(s1): \n");
printf("sizeof(s1) = %d\n", sizeof(s1));
printf("strlen(s1) = %d\n", strlen(s1));
printf("\nbefore swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);
printf("\nswapping...\n");
strcpy(tmp, s1);
strcpy(s1, s2);
strcpy(s2, tmp);
printf("\nafter swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);
return 0;
}
2
1 #include <stdio.h>
2 #include <string.h>
3 #define N 80
4 int main()
5 {
6 char *s1 = "Learning makes me happy";
7 char *s2 = "Learning makes me sleepy";
8 char *tmp;
9 printf("sizeof(s1) vs. strlen(s1): \n");
10 printf("sizeof(s1) = %d\n", sizeof(s1));
11 printf("strlen(s1) = %d\n", strlen(s1));
12 printf("\nbefore swap: \n");
13 printf("s1: %s\n", s1);
14 printf("s2: %s\n", s2);
15 printf("\nswapping...\n");
16 tmp = s1;
17 s1 = s2;
18 s2 = tmp;
19 printf("\nafter swap: \n");
20 printf("s1: %s\n", s1);
21 printf("s2: %s\n", s2);
22 return 0;
23 }
实验任务3
#include <stdio.h> void str_cpy(char *target, const char *source); void str_cat(char *str1, char *str2); int main() { char s1[80], s2[20] = "1984"; str_cpy(s1, s2); puts(s1); str_cat(s1, " Animal Farm"); puts(s1); return 0; } void str_cpy(char *target, const char *source) { while (*target++ = *source++) ; } void str_cat(char *str1, char *str2) { while (*str1) str1++; while (*str1++ = *str2++) ; }
试验任务4
#include <stdio.h>
#define N 80
int func(char *);
int main()
{
char str[80];
while (gets(str) != NULL)
{
if (func(str))
printf("yes\n");
else
printf("no\n");
}
return 0;
}
int func(char *str)
{
char *begin, *end;
begin = end = str;
while (*end)
end++;
end--;
while (begin < end)
{
if (*begin != *end)
return 0;
else
{
begin++;
end--;
}
}
return 1;
}
试验任务5
#include <stdio.h>
#define N 80
void func(char *);
int main()
{
char s[N];
while (scanf("%s", s) != EOF)
{
func(s);
puts(s);
}
return 0;
}
void func(char *str)
{
int i;
char *p1, *p2, *p;
p1 = str;
while (*p1 == '*')
p1++;
p2 = str;
while (*p2)
p2++;
p2--;
while (*p2 == '*')
p2--;
p = str;
i = 0;
while (p < p1)
{
str[i] = *p;
p++;
i++;
}
while (p <= p2)
{
if (*p != '*')
{
str[i] = *p;
i++;
} p
++;
}
while (*p != '\0')
{
str[i] = *p;
p++;
i++;
}
str[i] = '\0';
}
试验任务6
1
#include <stdio.h>
#include <string.h>
void sort(char *name[], int n);
int main()
{
char *course[4] = {"C Program",
"C++ Object Oriented Program",
"Operating System",
"Data Structure and Algorithms"};
int i;
sort(course, 4);
for (i = 0; i < 4; i++)
printf("%s\n", course[i]);
return 0;
}
void sort(char *name[], int n)
{
int i, j;
char *tmp;
for (i = 0; i < n - 1; ++i)
for (j = 0; j < n - 1 - i; ++j)
if (strcmp(name[j], name[j + 1]) > 0)
{
tmp = name[j];
name[j] = name[j + 1];
name[j + 1] = tmp;
}
}
2
#include <stdio.h>
#include <string.h>
void sort(char *name[], int n);
int main()
{
char *course[4] = {"C Program",
"C++ Object Oriented Program",
"Operating System",
"Data Structure and Algorithms"};
int i;
sort(course, 4);
for (i = 0; i < 4; i++)
printf("%s\n", course[i]);
return 0;
}
void sort(char *name[], int n)
{
int i, j, k;
char *tmp;
for (i = 0; i < n - 1; i++)
{
k = i;
for (j = i + 1; j < n; j++)
if (strcmp(name[j], name[k]) < 0)
k = j;
if (k != i)
{
tmp = name[i];
name[i] = name[k];
name[k] = tmp;
}
}
}
试验任务7
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 5
int check_id(char *str);
int main(){
char *pid[N]={"31010120000721656X",
"330106199609203301",
"53010220051126571",
"510104199211197977",
"53010220051126133Y"};
int i;
for(i=0;i<N;++i)
if(check_id(pid[i]))
printf("%s\tTure\n",pid[i]);
else
printf("%s\tFalse\n",pid[i]);
system("pause");
return 0;
}
int check_id(char *str){
int n,j,k;
char bj[]="1234567890X";
for(n=0;(str[n])!='\0';n++)
;
j=0;
for(k=0;k<=10;k++){
if(str[n-1]==bj[k])
j=1;
}
if((j==1)&&(n==18))
return 1;
else
return 0;
}
实验任务8
#include <stdio.h>
#include <stdlib.h>
#define N 80
void encoder(char *s);
void decoder(char *s);
int main(){
char words[N];
printf("输入英文文本：");
while(gets(words)!=NULL){
printf("编码后的英文文本：");
encoder(words);
printf("%s\n",words);
printf("对编码后的英文文本解码：");
decoder(words);
printf("%s\n",words);
printf("输入英文文本：");
}
system("pause");
return 0;
}
void encoder(char *s){
int i,j;
for(i=0;s[i]!='\0';i++)
;
i--;
for(j=0;j<=i;j++){
if(((s[j]>=65)&&(s[j]<=89))||((s[j]>=97)&&(s[j]<=121)))
s[j]=s[j]+1;
else{
if(s[j]==90)
s[j]=65;
else{
if(s[j]==122)
s[j]=97;
else
;
}
}
}
}
void decoder(char *s){
int i,j;
for(i=0;s[i]!='\0';i++)
;
i--;
for(j=0;j<=i;j++){
if(((s[j]>=66)&&(s[j]<=90))||((s[j]>=98)&&(s[j]<=122)))
s[j]=s[j]-1;
else{
if(s[j]==65)
s[j]=90;
else{
if(s[j]==97)
s[j]=122;
else
;
}
}
}
}
标签：int,s1,char,++,实验,printf,include From： https://www.cnblogs.com/20040802h/p/17371795.html